Implicit differentiation help again

[A_Munk3y]

implicit differentiation help :) again!

Homework Statement

Use implicit differentiation
1) x/(y-x²)=1

and

2) (y²-1)³=x²-y

The Attempt at a Solution

1) x/(y-x²)=1
=> [(y-x²)(1)-(x)((1*dy/dx)-2x)]/[(y-x)²]²=0
=> y-x²-x(dy/dx)+2x²=[(y-x)²]²

I think I'm going to stop here because I'm pretty sure it's wrong so no reason to keep putting what i got. What am i doing wrong here?2) (y²-1)³=x²-y
=> 3(y²-1)²*(2y(dy/dx) = 2x-(dy/dx)
=> 3(y²-1)²*(2y(dy/dx))+(dy/dx) = 2x

Now here, i know the next step is
(dy/dx)(3(y²-1)²*(2y)+1)=2x

But i have no idea how i should get to this step. How did the dy/dx (the one that comes first) come out and where did the other dy/dx go (it became a 1??)
can anyone explain this to me?

 

[Mark44]

Homework Statement

Use implicit differentiation
1) x/(y-x²)=1

and

2) (y²-1)³=x²-y

The Attempt at a Solution

1) x/(y-x²)=1
=> [(y-x²)(1)-(x)((1*dy/dx)-2x)]/[(y-x)²]²=0

So far, so good.

=> y-x²-x(dy/dx)+2x²=[(y-x)²]²

Error above. When you multiply 0 by (y - x²)², you get 0. If you fix this, the resulting equation is fairly easy to solve for y'.

I think I'm going to stop here because I'm pretty sure it's wrong so no reason to keep putting what i got. What am i doing wrong here?


2) (y²-1)³=x²-y
=> 3(y²-1)²*(2y(dy/dx) = 2x-(dy/dx)
=> 3(y²-1)²*(2y(dy/dx))+(dy/dx) = 2x

This is [6y(y² -1) + 1]y' = 2x
The next step should be obvious.

Now here, i know the next step is
(dy/dx)(3(y²-1)²*(2y)+1)=2x

But i have no idea how i should get to this step. How did the dy/dx (the one that comes first) come out and where did the other dy/dx go (it became a 1??)
can anyone explain this to me?